From a group of 10 men and 5 women, four member committees are to be formed each of which must contain at…

Total number of ways $={}^{5}C_1\times {}^{10}C_3+{}^{5}C_2\times {}^{10}C_2+{}^{5}C_3\times {}^{10}C_1+{}^{5}C_4$

$=\frac{5!}{4!}\times \frac{10!}{7!3!}+\frac{5!}{2!3!}\times \frac{10!}{8!2!}+\frac{5!}{3!2!}\times \frac{10!}{9!}+\frac{5!}{4!}$

$=5\times \frac{10\times 9\times 8}{3\times 2}+\frac{5\times 4}{2\times 1}\times \frac{10\times 9}{2}+\frac{5\times 4}{2\times 1}\times 10+5$

$=5\times 5\times 3\times 8+5\times 2\times 5\times 9+10\times 10+5$

$=600+450+100+5=1155$

Total no. of ways $=1155$
No. of ways to form a committee having more women than men
$=5c_1\times 10c_{c_1}+5c_4=\frac{5!}{3!2!}\times \frac{10!}{9!}+\frac{5!}{4!}=10\times 10+5=105$

probability$=\frac{105}{1155}=\frac{1}{11}$